The cubes of natural nos. are grouped ad 1^3,(2^3,3^3),(4^3,5^3,5^3),........................, then the sum of nos. in nth group is

Lets first find out the first and last terms of the nth group.

The first terms are cubes of 1,2,4,7,....t_k

Sum of k terms = 1+2+4+7+....+t_k......................(1)

Sum of (k-1) terms = 1+2+4+7+....+t_k-1...............(2)

Subtracting (2) from (1),

Sum of k terms - Sum of (k-1) terms = 1+(2-1)+(4-2)+(7-4)+.....+(t_k-t_k-1) = 1+1+2+3+...upto k terms = 1+ (Sum of first (k-1) natural numbers

= 1 + (k(k-1))/2

But we also know that Sum of k terms - Sum of (k-1) terms is the kth term. So, kth term is 1 + (k(k-1))/2.

Now consider the last terms.

The last terms are First term + k.

=1 + (k(k-1))/2 + k

So, the nth set is [(1 + (n(n-1))/2)³, (1 + (n(n-1))/2 + 1)³,.....(1 + (n(n-1))/2+n)³]

For example, the fourth set is [(1 + (4(4-1))/2)³, (1 + (4(4-1))/2 + 1)³,.....(1 + (4(4-1))/2+4)³]

= [7³,8³,9³,10³]

Sum of terms of fourth set = Sum of cubes upto 10³ - Sum of cubes upto 6³

Similarly, Sum of terms of nth set = Sum of cubes upto (1 + (n(n-1))/2 + n)³ - Sum of cubes upto (1 + (n(n-1))/2 - 1 )³

 = Sum of cubes upto ((n²+n+2)/2)³ - Sum of cubes upto ((n²-n)/2 )³ 

Now apply the sum of cubes formulas to both these terms and simplify.
Sum of cubes upto ((n²+n+2)/2)³ = [ {(n²+n+2)/2} {(n²+n+2)/2 + 1}/2 ]²  

Sum of cubes upto ((n²-n)/2 )³ = [ {(n²-n)/2} {(n²-n)/2 + 1}/2 ]²

Then the final answer will be [ {(n²+n+2)/2} {(n²+n+2)/2 + 1}/2 ]² - [ {(n²-n)/2} {(n²-n)/2 + 1}/2 ]² 

(You will have to further simplify this)